13. A 145 g baseball is dropped from a tree 8m above the ground A) with what spe

Question

13. A 145 g baseball is dropped from a tree 8m above the ground A) with what speed would it hit the ground if air resistance could be ignored? B) if it actually hits the ground with a speed of 7.8 m/s, what is the magnitude of the average force of air resistance exerted on it? 13. A 145 g baseball is dropped from a tree 8m above the ground A) with what speed would it hit the ground if air resistance could be ignored? B) if it actually hits the ground with a speed of 7.8 m/s, what is the magnitude of the average force of air resistance exerted on it? A) with what speed would it hit the ground if air resistance could be ignored? B) if it actually hits the ground with a speed of 7.8 m/s, what is the magnitude of the average force of air resistance exerted on it?

Explanation / Answer

A) The velocity V under constant acceleration is V=gt
The height h is given in terms of time of free fall t as
h=0.5gt^2
since t=sqrt(2h/g)
then finally
V= g sqrt(2h/g)
V= sqrt(2gh)
V=sqrt(2 x 8 x 9.81)=12.52 m/s

B) F(total) = F(gravity) - F(air resistance)
F(gravity)=mg and then
F(air resistance)= F(gravity) - F(total)

Using the total acceleration a then
F(total) = ma
F(air resistance)=mg - ma=m(g-a)
Lets find a
still however using our acceleration a
V=sqrt(2ha) then a is

a= 0.5 V^2 / h finally

F(air resistance)=m( g - 0.5 V^2 / h )
F(air resistance)=.145 ( 9.81 - 0.5 (7.8)^2/8)
F(air resistance)= 0.871 N

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